Trying to write the bundle structure for $\Lambda^2(T^*(\mathbb P^3\times \mathbb P^3))$

Question

I'm trying to find an atlas and in general the bundle structure for $\Lambda^2(T^*(\mathbb P^3\times \mathbb P^3))$ respect to, for example, the local chart $(U_{01},\psi_{01})$, where $$U_{01}:=\{((x^0:x^1:x^2:x^3),(y^0:y^1:y^2:y^3))\in\mathbb P^3\times \mathbb P^3|x^0\ne 0,y^1\ne 0\}$$ and $\psi_{01}:U_{01}\subset\mathbb P^3\times \mathbb P^3\to\mathbb R^3\times \mathbb R^3$ is the standard chart that divides for the non-zero component.
In particular I was trying to build the bundle structure for $\Lambda^2(T^*(\mathbb P^3\times \mathbb P^3))$ starting from $T^*(\mathbb P^3\times \mathbb P^3)$ and I did this:

Could you tell me if it's correct?
How could I "extend" this idea to build a smooth structure for $\Lambda^2(T^*(\mathbb P^3\times \mathbb P^3))$?

Thank you in advance.

Answer 1

Question: "How could I "extend" this idea to build a smooth structure for Λ2(T∗(P3×P3))? Thank you in advance."

Answer: An "alternative" approach. Since your bundle is constructed using the pull back of bundles on projective 3-space there is another approach that is more indirect but simpler. It is easier to construct such a structure on $\mathbb{P}^3$ and then pull back this structure to your product $\mathbb{P}^3 \times \mathbb{P}^3$.

Let $X:=\mathbb{P}^3_{k}$ be real ($k:=\mathbb{R}$) projective 3-space and let $Y:=X \times_k X$. There is a formula for the cotagent bundle of a product:

$$\Omega^1_Y \cong p^*\Omega^1_X \oplus q^*\Omega^1_X .$$

There is an exact sequence

$$ 0 \rightarrow \Omega^1_X \rightarrow \mathcal{O}(-1)^{4} \rightarrow \mathcal{O}_X \rightarrow 0.$$

You seek a description of the exterior product

$$\wedge^2 (p^*\Omega_X \oplus q^*\Omega_X)$$

There is (Hartshorne, Ex.II.5.16) a filtration

$$0=F^3 \subseteq F^2 \subseteq F^1 \subseteq F^0:=\wedge^2 (\Omega^1_Y)$$

with

$$F^i/F^{i+1} \cong \wedge^i(p^*\Omega^1_X)\otimes \wedge^{2-i}(q^*\Omega^1_X).$$

This gives an isomorphism

$$C1.\text{ }\wedge^2(T^*_Y):=\wedge^2 (\Omega^1_Y) \cong $$

$$ p^*(\wedge^2 \Omega^1_X)\oplus q^*(\wedge^2 \Omega^1_X) \oplus p^*\Omega^1_X \otimes q^*\Omega^1_X $$

since pull back commutes with wedge products. Hence if you can construct charts for $\Omega^1_X$, you get charts for $\wedge^2(T^*_Y)$ using formula $C1$ and the projection maps $p,q$.

Exmaple: If $x_0,x_1$ are homogeneous coordinates on $C:=\mathbb{P}^1_k$ you get local coordinates $t:=x_1/x_0$ on $U_0$ and $s:=x_0/x_1:=1/t$ on $U_1$.

The cotangent module $\Omega^1_C$ is on $U_0$ a free module (a trivial bundle) on the basis $dt$. On the open set $U_1$ it has the basis $ds:=d(1/t)=-1/t^2dt$. Hence on the intersection $U_{01}:=U_0 \cap U_1$ you get

$$ds=-\frac{1}{t^2} dt:=\psi_{10}(t,s)dt.$$

The function $\psi_{10}(t,s)$ is invertible on $U_{01}$ and defines the coordinate transformation on this open set. Since $\Omega^1_{C/k}$ is an invertible sheaf, the above calculation proves that

$$ \Omega^1_{C/k} \cong \mathcal{O}(-2).$$

You may work with the real rank one vector bundle on $C$ or the locally trivial sheaf of rank one. The above construction uses the language of "locally trivial sheaves". There are projection maps

$$p,q: C \times C \rightarrow C$$

and you get the formula

$$\Omega^1_{C\times C/k} \cong p^*\mathcal{O}(-2)\oplus q^*\mathcal{O}(-2)$$

and

$$\wedge^1(\Omega^1_{C\times C/k}) \cong p^*\mathcal{O}(-2)\otimes q^*\mathcal{O}(-2).$$

Let $U_0 \times U_0 :=D(x_0)\times D(y_0) \subseteq C \times C$ have coordinates $t,s$ and let $U_0 \subseteq C$ have coorodinate $t$. Let $U_1\times U_0 \subseteq C \times C$ have coodinates $1/t,s$ and let $U_1 \subseteq C$ have coordinate $1/t$. It follows the pull back invertible sheaf/line bundle

$$p^*\Omega^1_C$$

is on $U_0\times U_0$ ( as a free module) the following module: $k[t,s]dt$

On the open set $U_1 \times U_0$ it is the free module $k[1/t,s]d(1/t)$.

The transition function is the same function $\psi_{10}$:

$$d(\frac{1}{t})=-\frac{1}{t^2}d(t):=\psi_{10}d(t).$$

When you "pull back" a vector bundle $E$ from $C$ to $C\times C$ via $p$, you get a vector bundle $p^*E$ on $C\times C$ of the same rank as $E$ with the "same transition functions".

Hence for the real projective line you can construct an atlas and transition functions if you understand the above constructions. Once you have understood this you get a similar calculation for $X$ and $Y$.

A problem with a direct approach is that you end up calculating a $15 \times 15$ matrix. With the indirect approach you calculate two $3\times 3$-matrices and one $9\times 9$-matrix.